Two sides of a Triangel have lengths of A and B, and the angle between them is Theta what value of Theta will maximixe the triagle's area? HINT
A=(1/2)AB sin Theta

so where do i start?

October 25th 2006, 05:33 PM

ThePerfectHacker

Quote:

Originally Posted by cyberdx16

Two sides of a Triangel have lengths of A and B, and the angle between them is Theta what value of Theta will maximixe the triagle's area? HINT
A=(1/2)xy sin Theta

so where do i start?

Actually you do not need derivatives.
When is, maximized?
When the sin of the angle is maximized, when is that?
When (that is largest value of sine function).
That happens when

October 25th 2006, 05:33 PM

topsquark

Quote:

Originally Posted by cyberdx16

Two sides of a Triangel have lengths of A and B, and the angle between them is Theta what value of Theta will maximixe the triagle's area? HINT
A=(1/2)AB sin Theta

so where do i start?

The maximum area will be where the derivative of the area function is 0. (Note: we also get minima from this so we need to prove that we actually have a maximum.)

A' = (1/2)AB cos(theta)

Setting this to zero we get
cos(theta) = 0

which happens at theta = pi/2, 3*pi/2 rad.

The 3*pi/2 rad angle is ridiculously large for a triangle, so use theta = pi/2 rad. Is this point a local maximum? You can either do the second derivative test (ie show that A''(pi/2) < 0 ) or simply look at the graph of A(theta) at pi/2. It turns out that this is a maximum for the function.